[PY3]——根据某个特定的字段来分组迭代访问一个字段或序列？/ itertools.groupby( )函数

Posted 2020-09-28 Jelly_lyj

问题

你有一个字典或者实例的序列，然后你想根据某个特定的字段(比如‘date’)来分组迭代访问。

 

解决方案

itertools.groupby( )函数

 

itertools.groupby(rows,key=itemgetter(‘字段‘))

• groupby( )函数扫描整个序列并且查找连续相同值（或者根据指定 key函数返回值相同）的元素序列
• 在调用groupby( )函数前，我们首先需要按照这个字段来排序（这和SQL语句中的group by的使用是一个道理）
• 因为groupby( )仅仅检查连续的元素，如果事先并没有排序完成的话，分组函数将得不到想要的结果

rows = [
    {‘address‘: ‘5412 N CLARK‘, ‘date‘: ‘07/01/2012‘},
    {‘address‘: ‘5148 N CLARK‘, ‘date‘: ‘07/04/2012‘},
    {‘address‘: ‘5800 E 58TH‘, ‘date‘: ‘07/02/2012‘},
    {‘address‘: ‘2122 N CLARK‘, ‘date‘: ‘07/03/2012‘},
    {‘address‘: ‘5645 N RAVENSWOOD‘, ‘date‘: ‘07/02/2012‘},
    {‘address‘: ‘1060 W ADDISON‘, ‘date‘: ‘07/02/2012‘},
    {‘address‘: ‘4801 N BROADWAY‘, ‘date‘: ‘07/01/2012‘},
    {‘address‘: ‘1039 W GRANVILLE‘, ‘date‘: ‘07/04/2012‘},
]
from operator import itemgetter
from itertools import groupby
rows.sort(key=itemgetter(‘date‘))
print(rows)

　　[{‘date‘: ‘07/01/2012‘, ‘address‘: ‘5412 N CLARK‘}, {‘date‘: ‘07/01/2012‘, ‘address‘: ‘4801 N BROADWAY‘}, {‘date‘: ‘07/02/2012‘, ‘address‘: ‘5800 E 58TH‘}, {‘date‘: ‘07/02/2012‘, ‘address‘: ‘5645 N RAVENSWOOD‘}, {‘date‘: ‘07/02/2012‘, ‘address‘: ‘1060 W ADDISON‘}, {‘date‘: ‘07/03/2012‘, ‘address‘: ‘2122 N CLARK‘}, {‘date‘: ‘07/04/2012‘, ‘address‘: ‘5148 N CLARK‘}, {‘date‘: ‘07/04/2012‘, ‘address‘: ‘1039 W GRANVILLE‘}]


for date,items in groupby(rows,key=itemgetter(‘date‘)):
    print(date)
    for i in items:
        print(‘ ‘,i)
07/01/2012
{‘date‘: ‘07/01/2012‘, ‘address‘: ‘5412 N CLARK‘}
{‘date‘: ‘07/01/2012‘, ‘address‘: ‘4801 N BROADWAY‘}
07/02/2012
{‘date‘: ‘07/02/2012‘, ‘address‘: ‘5800 E 58TH‘}
{‘date‘: ‘07/02/2012‘, ‘address‘: ‘5645 N RAVENSWOOD‘}
{‘date‘: ‘07/02/2012‘, ‘address‘: ‘1060 W ADDISON‘}
07/03/2012
{‘date‘: ‘07/03/2012‘, ‘address‘: ‘2122 N CLARK‘}
07/04/2012
{‘date‘: ‘07/04/2012‘, ‘address‘: ‘5148 N CLARK‘}
{‘date‘: ‘07/04/2012‘, ‘address‘: ‘1039 W GRANVILLE‘}